As local martingales are semimartingales, they have a well-defined quadratic variation. These satisfy several useful and well known properties, such as the Ito isometry, which are the subject of this post. First, the covariation [*X*,*Y*] allows the product *XY* of local martingales to be decomposed into local martingale and FV terms. Consider, for example, a standard Brownian motion *B*. This has quadratic variation and it is easily checked that is a martingale.

Lemma 1IfXandYare local martingales thenXY-[X,Y] is a local martingale.

In particular, is a local martingale for all local martingalesX.

*Proof:* Integration by parts gives

which, by preservation of the local martingale property, is a local martingale.

For square integrable martingales, it is possible to drop the `local’ from the statement above. That is, will be a proper martingale. Before proving this, the following simple inequality will be useful. This is actually a special case of the much more general *Burkholder-Davis-Gundy inequalities*, which will be covered in a later post. Recall that denotes the maximum process of *X*.

Lemma 2LetXbe a local martingale with . Then,

(1)

*Proof:* By Lemma 1 there exist stopping times such that the stopped processes and are martingales. So, . Also, by Doob’s inequality, giving

Letting *n* increase to infinity and applying monotone convergence gives (1).

With the aid of this inequality, we can now prove that for square integrable martingales, will be a proper martingale. In fact, we obtain a necessary and sufficient condition for any local martingale to be a square integrable martingale.

Lemma 3A local martingaleXis a square integrable martingale if and only if and [X] is integrable, in which case is a martingale.

*Proof:* Replacing *X* by , we may suppose that . If [*X*] is integrable then Lemma 2 gives , so is a local martingale (by Lemma 1) and dominated by the integrable random variable (for ). Hence, it is a proper martingale. Conversely, suppose that *X* is a square integrable martingale. Then, Doobs inequality shows that is finite and, by Lemma 2, [*X*] is integrable.

In one direction, the above lemma generalizes in the obvious way to products of square integrable martingales.

Lemma 4LetXandYbe cadlag square integrable martingales. Then, [X,Y] has integrable variation over any bounded time interval, andXY-[X,Y] is a martingale.

*Proof:* By the polarization identity, . So, [*X*,*Y*] has variation bounded by over any interval [0,*t*] which, by Lemma 2, is integrable. Then

is a martingale, again using Lemma 2.

Recall that quadratic variations of stochastic integrals satisfy the following fundamental identity

Putting this together with the results above gives the *Ito Isometry*. In the following theorem, the random variable has zero mean, by the martingale property, and (2) gives a simple expression for the variance. Historically, the Ito isometry was first established for a Brownian motion *B* in which case it reads,

Equation (2) represents an extension to more general local martingales.

Theorem 5 (Ito Isometry)LetXbe a local martingale and be a predictable process such that has finite expectation. Then, isX-integrable, is an -bounded martingale and

(2)

By martingale convergence, the limit of the -bounded martingale *M* exists almost surely. So, there is no problem with integrating up to infinity in (2). Also, replacing by for any stopping time shows that (2) still holds if instead we integrate up to an arbitrary stopping time.

*Proof:* Let us first assume that is *X*-integrable and set . Then, is integrable. So is locally integrable and, therefore, *M* is locally square integrable. By preservation of the local martingale property, *M* is a local martingale. By (1),

so *M* is a uniformly -bounded martingale and Lemma 3 says that is a martingale giving,

as required.

Now, suppose that is any predictable process such that is integrable. It just needs to be shown that if is a sequence of bounded predictable processes tending to zero then tends to zero in probability, so that is X-integrable. However, by the above argument,

By dominated convergence for Lebesgue-Stieltjes integration, tends to zero, and is dominated by *U*. Then dominated convergence for sequences of random variables says that , so tends to zero in and, therefore, also in probability.

More generally, the Ito isometry can be used to express the covariance between two stochastic integrals.

Theorem 6LetX,Ybe local martingales and be predictable processes such that and are integrable. Then, are respectivelyX-integrable andY-integrable, has integrable variation and,

(3)

*Proof:* Theorem 5 says that is *X*-integrable, is *Y* integrable and , are -bounded martingales. The Kunita-Watanabe inequality shows that the variation of is bounded by

which is integrable. Furthermore, by Lemma 4, is a martingale and

giving (3).

We now show that for any local martingale *X* and predictable process , local integrability of is sufficient to guarantee that is *X*-integrable. In fact, it is enough that is locally integrable, which is a weaker condition, but the proof of that will have to wait until we have covered the Burkholder-Davis-Gundy inequalities.

Lemma 7For any local martingaleXand predictable process , the following are equivalent.

- is
X-integrable and .- is locally integrable.

*Proof:* Suppose that the first property is satisfied. Then, is locally square integrable. So, is locally integrable showing that is locally integrable.

Conversely, suppose that 2 is satisfied. Then, there are stopping times such that have finite expectation. So Theorem 5 says that are *X*-integrable. As *X*-integrability is a local property, this shows that is *X*-integrable and, by preservation of the local martingale property, is a local martingale. Finally, is locally integrable, so .

It is not true that, for a local martingale *X* and predictable process , finiteness of at all times implies that is *X*-integrable. Consider the following counter-example. Let be an exponentially distributed random variable, so , and set . This is a martingale under its natural filtration , and its quadratic variation is . If then is finite. However, cannot be *X*-integrable, because, that would lead to

whenever , which diverges as *t* decreases to zero.

It is true on the other hand, that finiteness of implies *X*-integrability for *continuous* local martingales, since continuous processes are locally bounded and Lemma 7 can be applied.

Finally, note that Lemma 1 can be strengthened to uniquely determine the covariation of local martingales. In fact, this is sometimes used as the initial definition of quadratic covariations.

Lemma 8LetX,Ybe local martingales. Then, [X,Y] is theuniqueFV process starting at 0 and satisfying the following.

XY-[X,Y] is a local martingale.- .

*Proof:* That [*X*,*Y*] satisfies property 1 is given by Lemma 1, and 2 is a basic property of the covariation. Conversely, suppose that *V* is any other FV process starting at zero and satisfying properties 1 and 2. Then will be a continuous local martingale. As it is a continuous FV process, *W* will have zero quadratic variation. So Lemma 3 says that is a martingale, giving , and V=[*X*,*Y*].

Dear Almost Sure,

thank you very much for your notes. I have a question. If we consider the Ito’s Isometry for the Brownian Motion then we will have expected value of squared stochastic integral is equal to the expected value of the integral of the squared function. Is the latter integral in the Riemann or Lebesgue sense?

Thank you.

Sincerely,

Alex.

Comment by Alex — 1 May 11 @ 3:45 PM |

The latter integral is in the Lebesgue sense — which is equivalent to the Riemann sense if the paths of are Riemann integrable (but that is not a requirement).

Comment by George Lowther — 2 May 11 @ 12:57 AM |

Dear Almost Sure,

I was reading your notes on stochastic calculus and I was wondering if there is an expression for the Riemann integral (with respect to t) of a Poisson Process or Compound Poisson Process? What is this integral equal to?

Is it also true (by Fubini Theorem) that the expectation of this integral is equal to the integral of the expectation?

Thank you.

Sincerely,

Alex.

Comment by Alex — 12 October 11 @ 5:50 PM |